2)
q= 16 kJ
m of brick = 938 g
T1 = 19.5o C
T2 = 35o C
∆T = T2 - T1

35 – 19.5 o C

15.5o Cq = mc∆T
c = q/ m∆T
c = 16kJ / (938g X 15.5o C)
c = 1.1 J/g oC

4)
m of kettle = 2.0 Kg
c = 0.385J/g.oC
T1 = 20.0 C
T2 = 80 C
∆T = 60 C
q (kettle) = mc∆T
q =2.0 kg X (0.385 J / g C) X 60 C
q = 46.2 kJ
q (water) = mc∆T
= 0.5 kg X (4.18 J/g C) X 60 C
= 125.4 kJ
Total heat needed to raise the temperature of the kettle and its content 46.2 kJ + 125.4 kJ = 172 kJ

5)
200.0 L H2O (l)
ΔT of H2O = 45°C
Δ H of C3H8 = -2220 kJ/mol converts to -2.22 x 10^6 J/mol
molar mass C3H8 = 44 g/mol
molar mass H2O = 18 g/mol
mass C3H8 = ?

V H2O --> m H2O
m = 200 L X 1000 mL/L X 1 g/1 mL
= 200,000 g

mol C3H8 = (m X c X ΔT)/ ΔH
= (200,000 g X 4.18 J/g*°C X 45°C)/ 2.22 x 10^6 J/mol
= 16.9 mol

mol C3H8 --> m C3H8
m = 16.9 mol X 44g/mol
m = 743.6 g

(textbook gives 547 g, but I think it’s wrong)
(I think the text is wrong too... I got 754.3g propane -Morgan)
This question makes no sense as it was done: hang on ... correctly, break it down into steps:
The energy gained by the water is:
Quantity equals 200 kg X 4200 J/kgC X 45 C equals 3.78 X 10 E 7 Joules
Now, how much energy did the propane supply? The same amount (but negative = -3.78 X 10 E 7 Joules)
How much propane was needed? Well we know that each mole of propane provides 2.2 X 10 E 6 Joules ... (again negative because it is exothermic.) So how many moles of propane are necessary:
-3.78 X 10 E 7 / -2.2 X 10 E 6 = 17.18 moles
Each mole of propane has a mass of 44g
17.18 moles X 44g / mole = 755.92g .... so your answer is correct, but I couldn't follow your logic ... Paul.


10)
C6H12O6 + 6O2 -> 6CO2 + 6H2O + 2813 kJ
C2H6 + 7/2O2 -> 2CO2 + 3H2O + 1369 kJ

C6H12O6 + 6O2 -> 6CO2 + 6H2O ∆H= - 2813 kJ
3H2O + 2CO2 -> 7/2O2 + C2H6 ∆H= + 1369

C6H12O6 + 5/2O2 -> 4CO2 + 3H2O + C2H6 + 1454 kJ ∆H= - 1454 kJ

Molar mass of glucose: 6(12) + 12 (1) + 6(16) = 180 g/mol

500g(1mol/180g) = 2.78 mol
2.78mol ( -1454 kJ/mol) = -4830.89 kJ


11)
∆T=81.8 C
c= 4.18 kJ/kg C
m= 3.77 kg
∆H= -802 kJ/mol
Molar Mass of Methane= 16 g/mol

Q= mc∆T
Q= 3.77 kg X 4.18 kJ/kg C X 81.8 C
Q= 1289 kJ

number of moles of methane Q/∆H 11 289 kJ/ 802 kJ per mol = 1.6 mol

Mass of methane needed # of moles X Molar Mass 1.6 mol X 16 g/mol = 25.7g

Therefore, the mass of methane needed is 25.7 g.

12)
CH3COCOOH → CH3COOH + CO
∆H =?
Reference Equations:
a) CH3COCOOH + 5/2 O2 → 3CO2 + 2H2O ∆H =-1275 kJ/mol
b) CH3COOH + 2O2 → 2CO2 + 2H2O ∆H =-875.3 kJ/mol
c) CO + ½ O2 → CO2 ∆H =-282.7 kJ/mol
Solving:
CH3COCOOH + 5/2 O2 → 3CO2 + 2H2O ∆H = -1275 kJ/mol
(remains same)
2CO2 + 2H2O → CH3COOH + 2O2 ∆H= -1(-875.3 kJ/mol) 875.3 kJ
(multiply equation by -1 as equation is reversed)
CO2 → CO + ½ O2 ∆H = -1(-282.7 kJ/mol) 282.7 kJ
(equation is reversed so multiply by -1)
Answer:
CH3COCOOH → CH3COOH + CO
∆H = -117 kJ/mol